Answer
$\lim\limits_{x \to \ -2}\frac{x^2}{x-3}=\frac{(-2)^2}{-2-3}=\frac{4}{-5}=-0.8$
Work Step by Step
If we want to calculate the value of $\lim\limits_{x \to \ -2}\frac{x^2}{x-3}$,
we can use the formula of:
$\lim \limits_{x\to a}f(x)=f(a)$, here it means:
$\lim\limits_{x \to \ -2}\frac{x^2}{x-3}=\frac{(-2)^2}{-2-3}=\frac{4}{-5}=-0.8$
